1. Exponential and Logarithmic Equations LESSON 3–4

2. Key Concept 1 A. Solve 4 x + 2 = 16 x – 3. 4 x + 2 = 16 x – 3 Original equation 4 x + 2 = (4 2 ) x – 3 4 2 = 16 4 x + 2 = 4 2x – 6 Power of a Power x + 2= 2x – 6One-to-One Property 2= x – 6Subtract x from each side. 8= xAdd 6 to each side. Answer: 8

3. Example 1 Solve Exponential Equations Using One-to- One Property B. Solve. Original equation Power of a Power One-to-One Property.

4. Example 1 A.1 B. C.2 D.–2 Solve 25 x + 2 = 5 4x.

5. Example 2 Solve Logarithmic Equations Using One-to- One Property A. Solve 2 ln x = 18. Round to the nearest hundredth. Method 1 Use exponentiation. 2 ln x= 18 Original equation ln x= 9 Divide each side by 2. e ln x = e 9 Exponentiate each side. x= e 9 Inverse Property x≈ 8103.08 Use a calculator. Method 2 Write in exponential form. 2 ln x= 18 Original equation ln x= 9 Divide each side by 2. x= e 9 Write in exponential form. x≈ 8103.08 Use a calculator. Answer: 8103.08

6. Example 2 Solve Logarithmic Equations Using One-to- One Property B. Solve 7 – 3 log 10x = 13. 7 – 3 log 10x= 13Original equation –3 log 10x= 6Subtract 7 from each side. log 10x= –2Divide each side by –3. 10 –2 =10xWrite in exponential form. 10 –3 = xDivide each side by 10. = x = 10 –3. Answer:

7. Example 2 Solve 2 log 2 x 3 = 18. A.81 B.27 C.9 D.8

8. Key Concept 2 A. Solve log 2 5 = log 2 10 – log 2 (x – 4). log 2 5= log 2 10 – log 2 (x – 4)Original equation log 2 5=Quotient Property 5=One-to-One Property 5x – 20= 10Multiply each side by x – 4.5 5x= 30Add 20 to each side. x= 6Divide each side by 5. Answer: 6

9. Example 3 Solve Exponential Equations Using One-to- One Property B. Solve log 5 (x 2 + x) = log 5 20. log 5 (x 2 + x) = log 5 20Original equation x 2 + x= 20One-to-One Property x 2 + x – 20= 0Subtract 20 from each side. (x – 4)(x + 5)= 0Factor x 2 + x – 20 into linear factors. x= –5 or 4Solve for x. Check this solution. Answer: –5, 4

10. Example 3 Solve log 3 15 = log 3 x + log 3 (x – 2). A.5 B.–3 C.–3, 5 D.no solution

11. Example 4 Solve Exponential Equations A. Solve 3 x = 7. Round to the nearest hundredth. 3 x = 7Original equation log 3 x = log 7Take the common logarithm of each side. x log 3= log 7Power Property x= or about 1.77Divide each side by log 3 and use a calculator. Answer:1.77 B. Solve e 2x + 1 = 8. Round to the nearest hundredth. e 2x + 1 = 8Original equation ln e 2x + 1 = ln 8Take the natural logarithm of each side. 2x + 1= ln 8Inverse Property x= or about 0.54Solve for x and use a calculator. Answer:0.54

12. Example 4 Solve 4 x = 9. Round to the nearest hundredth. A.0.63 B.1.58 C.2.25 D.0.44

13. Example 5 Solve in Logarithmic Terms Solve 3 6x – 3 = 2 4 – 4x. Round to the nearest hundredth. 3 6x – 3 = 2 4 – 4x Original equation ln 3 6x – 3 = ln 2 4 – 4x Take the natural logarithm of each side. (6x – 3) ln 3= (4 – 4x) ln 2Power Property 6x ln 3 – 3 ln3= 4 ln 2 – 4x ln 2Distributive Property 6x ln 3 + 4x ln 2= 4 ln 2 + 3 ln3Isolate the variable on the left side of the equation. x(6 ln 3 + 4 ln 2)= 4 ln 2 + 3 ln3Distributive Property (could just divide here… hard to type in)

14. Example 5 Solve in Logarithmic Terms x(ln 3 6 + ln 2 4 )= ln 2 4 + ln 3 3 Power Property x ln [3 6 (2 4 )]= ln [2 4 (3 3 )]Product Property x ln 11,664= ln 4323 6 (2 4 ) = 11,664 and 2 4 (3 3 ) = 432 x= Divide each side by ln 11,664. Answer:0.65 x≈ 0.65Use a calculator.

15. Example 5 Solve 4 x + 2 = 3 2 – x. Round to the nearest hundredth. A.1.29 B.1.08 C.0.68 D.–0.23

16. Example 6 Solve e 2x – e x – 2 = 0. e 2x – e x – 2= 0Original equation u 2 – u – 2= 0Write in quadratic form by letting u = e x. (u – 2)(u + 1)= 0Factor. u= 2 or u= –1Zero Product Property e x = 2e x = –1Replace u with e x. Solve Exponential Equations in Quadratic Form ln e x = ln 2ln e x = ln (–1)Take the natural logarithm of each side. x= ln 2 x= ln (–1) Inverse Property or about 0.69 The only solution is x = ln 2 because ln (–1) is extraneous. Answer: 0.69

17. Example 6 Solve e 2x + e x – 12 = 0. A.ln 3 B.ln 3, ln 4 C.ln 4 D.ln 3, ln (–4)

18. Example 7 Solve Logarithmic Equations Solve log x + log (x – 3) = log 28. log x + log (x – 3)= log 28Original equation log x(x – 3)= log 28Product Property log (x 2 – 3x)= log 28Simplify. x 2 – 3x= 28One-to-One Property x 2 – 3x – 28= 0Subtract 28 from each side. (x – 7)(x + 4)= 0Factor. x = 7 or x= – 4Zero Product Property The only solution is x = 7 because –4 is an extraneous solution. Answer: 7

19. Example 8 Solve log (3x – 4) = 1 + log (2x + 3). log (3x – 4)= 1 + log (2x + 3) Check for Extraneous Solutions log (3x – 4) – log (2x + 3)= 1 = 1 = log 10 1 = log 10 = 10 3x – 4= 10(2x + 3) 3x – 4= 20x + 30 –17x= 34 x= –2 But wait… must check for extraneous solutions

20. Example 8 Answer:no solution Check for Extraneous Solutions Check log (3x – 4)= 1 + log (2x + 3) log (3(–2) – 4)= 1 + log (2(–2) + 3) log (–10)= 1 = log (–1) Since neither log (–10) nor log (–1) is defined, x = –2 is an extraneous solution.

21. Example 9 Model Exponential Growth A. CELL PHONES This table shows the number of cell phones a new store sold in March and August of the same year. If the number of phones sold per month is increasing at an exponential rate, identify the continuous rate of growth. Then write the exponential equation to model this situation.

22. Example 9 Model Exponential Growth Let N(t) represent the number of cell phones sold at the end of t months and assume continuous growth. Then the initial number N 0 is 88 cell phones sold and the number of cell phones sold N after a time of 5 months, the number of months from March to August, is 177. Use this information to find the continuous growth rate k. N (t)= N 0 e kt Exponential Growth Formula 177= 88e 5k N(5) = 177, N 0 = 88, and t = 5 = e 5k Divide each side by 88.

23. Example 9 Answer:13.98%; N (t) = 88e 0.1398t Model Exponential Growth = ln e 5k Take the natural logarithm of each side. = 5kInverse Property = kDivide each side by 5 0.1398≈ kUse a calculator.

24. Example 9 Model Exponential Growth B. CELL PHONES Use your model to predict the number of months it will take for the store to sell 500 phones in one month. N (t) = 88e 0.1398t 500= 88e 0.1398t = ln e 0.1398t = e 0.1398t = 0.1398t = t 12.4≈ t According to this model, the store will sell 500 phones in a month in about 12.4 months. Answer:12.4 months